The final temperature = 36 °C
<h3>Further explanation</h3>The balanced combustion reaction for C₆H₆
2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ
MW C₆H₆ : 78.11 g/mol
mol C₆H₆ :
Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol
Heat transferred to water :
Q=m.c.ΔT
Answer:- 0.800 moles of the gas were collected.
Solution:- Volume, temperature and pressure is given for the gas and asks to calculate the moles of the gas.
It is an ideal gas law based problem. Ideal gas law equation is used to solve this. The equation is:
PV=nRT
Since it asks to calculate the moles that is n, so let's rearrange this for n:
V = 19.4 L
T = 17 + 273 = 290 K
P = 746 mmHg
we need to convert the pressure from mmHg to atm and for this we divide by 760 since, 1 atm = 760 mmHg
P = 0.982 atm
R =
Let's plug in the values in the equation to get the moles.
n = 0.800 moles
So, 0.800 moles of the gas were collected.
1) Balance the chemical equation.
2) List the known and unknown quantities.
Reactant 1: Ca(NO3)2.
Amount of substance: 3.40 mol.
Reactant 2: Li3PO4.
Amount of substance: 2.40 mol.
Product: Ca3(PO4)2
Mass: unknown.
3) Which is the limiting reactant?
<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.
<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.
4) Moles of Ca3(PO4)2 produced from the limiting reactant.
We have 3.40 mol Ca(NO3)2 of the limiting reactant.
The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.
5) Mass of Ca3(PO4)2 produced.
The molar mass of Ca3(PO4)2 is 310.1767 g/mol.
<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.
Option D.
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